Clocks are oscillators even atomic clocks.
Basically some object travels back and forth between 2 points in space say A and B
One traversal of space from A to B and back to A is made the unit of time.
Now lets take the path from A to B.
Put point A lower in a gravitational potential than point B (parallel to the direction of force).
Let the oscillator go.
The motion from A to B is going to slow because it is experiencing a decelerating gravity force.
B to A transit will be faster because of an accelerating gravity force.
Rotate the oscillation path 90 degrees and the oscillation time will change.
Now lets start moving and rotating this oscillator around in the gravity field that falls off at the square of the distance perpendicular to a curved sphere.
The unit of time which is a derivative of the A to B - B to A motion will keep
changing.
The only way for A to B motion and the B to A motion to experience an equal force across the entire path of motion is thus,
A path of motion in which the gravitational force is always equal.
For a spherical object of uniform density, that path curved.
The path is always perpendicular to the gradient of force.
To put is simply.
The PATH (A-B and B-A) of the oscillator MUST stay at a constant height above the surface of the sphere.
The surface of a sphere is curved so a constant height is curved.
Like a said good luck finding a oscillator the follows a curved path parallel to the surface of a sphere.
The dashed line in the image is a Gaussian sphere. The gravitational force across the surface of the imaginary sphere is uniform according to Newton.
This is because force falls of linear with distance i.e. inverse square law.
Relativity on the other hand is non-linear. A uniform path within the field can not be computed because initial values of 2 masses (their mass and field strength) can not computed.
Basically some object travels back and forth between 2 points in space say A and B
One traversal of space from A to B and back to A is made the unit of time.
Now lets take the path from A to B.
Put point A lower in a gravitational potential than point B (parallel to the direction of force).
Let the oscillator go.
The motion from A to B is going to slow because it is experiencing a decelerating gravity force.
B to A transit will be faster because of an accelerating gravity force.
Rotate the oscillation path 90 degrees and the oscillation time will change.
Now lets start moving and rotating this oscillator around in the gravity field that falls off at the square of the distance perpendicular to a curved sphere.
The unit of time which is a derivative of the A to B - B to A motion will keep
changing.
The only way for A to B motion and the B to A motion to experience an equal force across the entire path of motion is thus,
A path of motion in which the gravitational force is always equal.
For a spherical object of uniform density, that path curved.
The path is always perpendicular to the gradient of force.
To put is simply.
The PATH (A-B and B-A) of the oscillator MUST stay at a constant height above the surface of the sphere.
The surface of a sphere is curved so a constant height is curved.
Like a said good luck finding a oscillator the follows a curved path parallel to the surface of a sphere.
The dashed line in the image is a Gaussian sphere. The gravitational force across the surface of the imaginary sphere is uniform according to Newton.
This is because force falls of linear with distance i.e. inverse square law.
Relativity on the other hand is non-linear. A uniform path within the field can not be computed because initial values of 2 masses (their mass and field strength) can not computed.
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